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If the position state $|x\rangle$ is complete and orthonormal I understand that $$\langle x | y \rangle = \delta(x-y).\tag1$$

However, why exactly $$\langle x' | x \rangle = \delta(x'-x)\tag2$$

How do we know that $x'$ is orthogonal to $x $?

here, they use eq. (2), but we don't know anything about $x'$.

And the explications here and here seems to always assume the statement above. Is it only something with the notation that I don't understand or is it something else?

Qmechanic
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InvisibleParticle
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3 Answers3

3

First,

$$ \langle x|p\rangle = e^{ipx} $$

as it is a plain wave. Now, with the identity

$$ \int\frac{dp}{2\pi}|p\rangle\langle p| = 1 $$ you can get $$ \langle x|x'\rangle = \int \frac{dp}{2\pi}\langle x|p\rangle\langle p|x'\rangle = \int \frac{dp}{2\pi}e^{ip(x-x')} = \delta(x-x') $$

Vicky
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Well in the momentum basis ($\hbar=1$):

$$ \langle x'|x \rangle \propto \int{e^{-ipx'}e^{ipx} dp} = \int e^{ip(x-x')} dp\propto \delta(x-x') $$

JEB
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$|x>$ is a state with definite position. It describe a state of a particle whose position is definitely at point x. Now in QM, the notion of inner products gives you how two states is connected with each other. Saying that the particle is definitely at point $x'$ is of course completely different in saying that the particle is at point x. That is why, that the inner product of $<x|x'> = 0 $ in the case that $x' \neq x$ and $<x|x'> = 1 $ if $x = x'$, under the assumption that the kets here are normalized states. In general, two states that are orthogonal means that they are completely different with each other.

King Meruem
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